Question

Ayuda por favor. Los necesito para hoy.

Gracias.

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Answer 1

Ejercicio 3

a) 1 - x³+ 2x - 3 (x²- 2x + 1)=

= 1 - x³ + 2x -3x² + 6x - 3 =

= -x³ - 3x² + 8x -2

b) (1 - x³ + 2x²) (x² - 2x + 1) =

=x² - 2x + 1 - $x^{5}$ + 2$x^{4}$ - x³ + 2$x^{4}$ - 4x³ + 2x²=

= -$x^{5}$ + 4$x^{4}$ - 5x³ + 3x² - 2x + 1

c) 3x² (1 - x³ + 2x²) - 2x³ (x² - 2x + 1)=

= 3x² - 3$x^{5}$ + 6$x^{4}$ - 2$x^{5}$ + 4$x^{4}$ - 2x³=

= -5$x^{5}$ + 10$x^{4}$ - 2x³ - 3x²

d) (x³ - 3x + 6) (x² - 2x + 1)=

= $x^{5}$ - 2$x^{4}$ + x³ - 3x³ + 6x² - 3x + 6x² - 12x + 6=

= $x^{5}$ - 2$x^{4}$ - 2x³ + 12x² - 15x + 6

Ejercicio 4

a) $\frac{x+1}{3} -\frac{x+4}{4} -\frac{3x}{5} =$

= $\frac{20(x+1)}{60} -\frac{15(x+4)}{60} -\frac{12 .3x}{60} =$

$=\frac{20x+20-15x-60-36x)}{60} =$

$=\frac{-31x-40}{60}$

b) $\frac{-2x}{6} -\frac{x+4}{4}-\frac{3}{2} =$

$=\frac{-4x}{12} -\frac{3(x+4)}{12} -\frac{18}{12} =$

$=\frac{-4x-3x-4-18}{12} =$

$=\frac{-7x-22}{12}$

c) $\frac{2(x+3)}{3} -\frac{x-1}{4} -\frac{3x}{5}=$

$=\frac{20(2x+6)}{60} -\frac{15(x-1)}{60} -\frac{12.3x}{60} =$

$=\frac{40x+120-15x+15-36x}{60} =$

$=\frac{-11x+135}{60}$